# Riemann-Lebesgue lemma applied

I am little stuck in the following:

I showed (with help) that:$$\frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+ 2\cos2m\theta$$

Now is the question: apply the Riemann-Lebesgue lemma on the following integral:

$$\int_{\pi/4}^{\pi/2}\frac{\sin(2m+1) \theta}{\sin \theta}$$

I know by definition that: $$\lim_{k \to \infty} \int_{a}^{b}f(x)\sin(kx) = 0$$ This looks a lot on the first line.So if I multiply this out, I obtain:

$$\lim_{k \to \infty} \int_{\pi/4}^{\pi/2}(\sin k \theta)(1 + 2 \cos2\theta + 2\cos4\theta+\dots+ 2\cos2m\theta) d\theta =0$$

using again the relation: $$2\cos r\theta\sin \theta=\sin(r\theta+\theta)-\sin(r\theta-\theta)$$

But to be honest, I don't know that I am really doing. When I multipy all this and integrate it should be zero? So am I proving the R-L lemma?

A second question, how do I get a series expansion about $$\pi/4$$, can I just apply it on the first line, using the standard procedure?

any help appreciated by me

I will give you some pointers: if you need to be more explicit, let me know. You don't need to prove the lemma as it is given to you. You have shown the sine term can be written as a sum of cosines ie. that they are equivalent. Substitute this sum into the integrand and then integrate over the limits. You should get $\pi/4$ minus a sum of terms. However, from the RL lemma and since
$f(\theta) = \frac{1}{\sin \theta}$
is piecewise continuous over the interval which is used for the limits of integration, this whole thing equals 0 by the RL lemma. Hence take the sum to the RHS and you have a series expansion for $\pi/4$.